P(t<X<t+dt)=f(t)dt ?

1. Sep 2, 2011

rukawakaede

Hi

Could anyone show to me that:

If dt is an infinitely small number, the probability that X is included within the interval (t, t + dt) is equal to f(t) dt ,i.e.
$Pr(t<X<t+dt) = f(t)dt$
where f is the probability density function.

this sentence is from wikipedia but I could not prove this to myself.

Thanks if anyone can help.

2. Sep 2, 2011

chiro

What are the limits of your integral? You can't really make sense of f(t)dt without some kind of constraint on the upper and lower limits.

3. Sep 2, 2011

rukawakaede

here dt is an infinitely small number. and i don't think the RHS is an integral.

maybe you can refer to http://en.wikipedia.org/wiki/Probability_density_function#Further_details

the last line of the Further details

Last edited: Sep 2, 2011
4. Sep 2, 2011

HallsofIvy

Staff Emeritus
The probabilty density function, f, is defined by the fact that
$$P(a< x< b)= \int_a^b f(x)dx$$
or, equivalently,
$$P(a< x< a+h)= \int_a^{a+h} f(x)dx$$

What Wikipedia gives is a "differential" form of that.

5. Sep 3, 2011

rukawakaede

I was trying to work out why
$$\int_a^{a+h} f(x)dx= f(a)h$$
when h infinitely small.

Could you tell me more about why this is true? or could you explain the differential form? I am still not fully understand yet.

6. Sep 3, 2011

Hurkyl

Staff Emeritus
First off, you almost surely do not mean much of what you said literally. You should spend some time thinking about what you actually meant

Anyways, if by "=" you meant "is approximately" and by "infinitely small" you meant "sufficiently small", it's true because of the mean value theorem and the definition of continuous function.

(P.S. if f is not assumed to be continuous, then the equation above is very false)

7. Sep 3, 2011

bpet

It's just a (dodgy) method for turning sums into integrals. In the long run it's worthwhile to learn Stieltjes integration so you can write probabilities and expectations as
$$E[g(X)] = \int_R g(x)dF(x)$$
which is valid whether or not the distribution F(x) has a density.